A ball falls under gravity from a height of 10m with an initiall downward velocity'' u''.it collides with the ground ,loses 50%of its kinetic energy in collission and then rises back to the same height. Find the initial velocity ''u'
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Answered by
106
Given :
height h=10m
Initial velocity = V°
let m be the mass of ball.
Ball is projected downward with velocity V° from height h.
Total energy =mgh +1/2mv°²
Total Energy collision = 50/100x mgh + 1/2 mv°²
the ball rises back to same height after collision hence
V°=√2gh
=√2x9.8x 10
=14 m/s
therefore initial velocity = 14m/s
height h=10m
Initial velocity = V°
let m be the mass of ball.
Ball is projected downward with velocity V° from height h.
Total energy =mgh +1/2mv°²
Total Energy collision = 50/100x mgh + 1/2 mv°²
the ball rises back to same height after collision hence
V°=√2gh
=√2x9.8x 10
=14 m/s
therefore initial velocity = 14m/s
Answered by
61
g = 9.8 m/s² and h = 10 m
velocity just before hitting the ground = v
v² = u² + 2 g h = u² + 20 g
KE = 1/2 m v²
KE after bouncing from the ground : 1/2 * m * (v/√2)²
So velocity after bouncing = v/√2
When the ball rises to height 10 m then:
0² = (v/√2)² - 2 g * 10
=> v² = 40 g
=> u² + 20 g = 40 g
=> u = √(20 g) = 14 m/sec
velocity just before hitting the ground = v
v² = u² + 2 g h = u² + 20 g
KE = 1/2 m v²
KE after bouncing from the ground : 1/2 * m * (v/√2)²
So velocity after bouncing = v/√2
When the ball rises to height 10 m then:
0² = (v/√2)² - 2 g * 10
=> v² = 40 g
=> u² + 20 g = 40 g
=> u = √(20 g) = 14 m/sec
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